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PHXI08:GRAVITATION

359719 An artificial satellite moving in a circular orbit around the earth has a total energy $E_{0}$ . Its potential energy is

1

- 2 E_{0}

2

E_{0}

3

- E_{0}

4

2 E_{0}

Explanation:

Potential energy $= 2 ($ total energy $) = 2 E_{0}$

PHXI08:GRAVITATION

359720 Match Column I with Column II. For a satellite in circular orbit, (where $M_{E}$ is the mass of the Earth, $m$ is mass of the satellite and $r$ is the radius of the orbit)
Column I
Column II
A
Kinetic energy
P
$- \frac{G M_{E} m}{2 r}$
B
Potential energy
Q
$\sqrt{\frac{G M_{E}}{r}}$
C
Total energy
R
$- \frac{G M_{E} m}{r}$
D
Orbital velocity
S
$\frac{G M_{E} m}{2 r}$

1 A - R, B - S, C - Q, D - P

2 A - Q, B - P, C - R, D - S

3 A - P, B - Q, C - S, D - R

4 A - S, B - R, C - P, D - Q

Explanation:

Kinetic energy $= \frac{G M_{E} m}{2 r}; A - S$
Potential energy $= - \frac{G M_{E} m}{r}; B - R$
Total energy $= - \frac{G M_{E} m}{2 r}; C - P$
Orbital velocity $= \sqrt{\frac{G M_{E}}{r}}; D - Q$
Option (4) is correct.

PHXI08:GRAVITATION

359721 The minimum energy required to launch a $m kg$ satellite from earth's surface in a circular orbit at an altitude of $2 R, R$ is the radius of earth, will be

1

3 mgR

2

\frac{5}{6} m g R

3

2 m g R

4

\frac{1}{5} m g R

Explanation:

Minimum required energy
$\begin{aligned} = T E_{f} - P E_{i} \\ = \frac{- G M m}{2 (R + 2 R)} - (\frac{- G M m}{R}) \\ = \frac{G M m}{R} (- \frac{1}{6} + 1) \\ = \frac{5}{6} \frac{G M m}{R} \\ = \frac{5}{6} \frac{g R^{2} \times m}{R} \\ = \frac{5}{6} m g R \end{aligned}$

PHXI08:GRAVITATION

359722 In a satellite if the time of revolution is $T$ , then kinetic energy is proportional to

1

T^{- 2 / 3}

2

\frac{1}{T}

3

\frac{1}{T^{2}}

4

\frac{1}{T^{3}}

Explanation:

The kinetic energy of the satellite is
$K = \frac{1}{2} m v_{0}^{2} = \frac{1}{2} m (\frac{G M}{r}) = \frac{G M m}{2 r}$
$\Rightarrow K α \frac{1}{r} α \frac{1}{T^{2 / 3}} α T^{- 2 / 3}$

PHXI08:GRAVITATION

359719 An artificial satellite moving in a circular orbit around the earth has a total energy $E_{0}$ . Its potential energy is

1

- 2 E_{0}

2

E_{0}

3

- E_{0}

4

2 E_{0}

Explanation:

Potential energy $= 2 ($ total energy $) = 2 E_{0}$

PHXI08:GRAVITATION

359720 Match Column I with Column II. For a satellite in circular orbit, (where $M_{E}$ is the mass of the Earth, $m$ is mass of the satellite and $r$ is the radius of the orbit)
Column I
Column II
A
Kinetic energy
P
$- \frac{G M_{E} m}{2 r}$
B
Potential energy
Q
$\sqrt{\frac{G M_{E}}{r}}$
C
Total energy
R
$- \frac{G M_{E} m}{r}$
D
Orbital velocity
S
$\frac{G M_{E} m}{2 r}$

1 A - R, B - S, C - Q, D - P

2 A - Q, B - P, C - R, D - S

3 A - P, B - Q, C - S, D - R

4 A - S, B - R, C - P, D - Q

Explanation:

Kinetic energy $= \frac{G M_{E} m}{2 r}; A - S$
Potential energy $= - \frac{G M_{E} m}{r}; B - R$
Total energy $= - \frac{G M_{E} m}{2 r}; C - P$
Orbital velocity $= \sqrt{\frac{G M_{E}}{r}}; D - Q$
Option (4) is correct.

PHXI08:GRAVITATION

359721 The minimum energy required to launch a $m kg$ satellite from earth's surface in a circular orbit at an altitude of $2 R, R$ is the radius of earth, will be

1

3 mgR

2

\frac{5}{6} m g R

3

2 m g R

4

\frac{1}{5} m g R

Explanation:

Minimum required energy
$\begin{aligned} = T E_{f} - P E_{i} \\ = \frac{- G M m}{2 (R + 2 R)} - (\frac{- G M m}{R}) \\ = \frac{G M m}{R} (- \frac{1}{6} + 1) \\ = \frac{5}{6} \frac{G M m}{R} \\ = \frac{5}{6} \frac{g R^{2} \times m}{R} \\ = \frac{5}{6} m g R \end{aligned}$

PHXI08:GRAVITATION

359722 In a satellite if the time of revolution is $T$ , then kinetic energy is proportional to

1

T^{- 2 / 3}

2

\frac{1}{T}

3

\frac{1}{T^{2}}

4

\frac{1}{T^{3}}

Explanation:

The kinetic energy of the satellite is
$K = \frac{1}{2} m v_{0}^{2} = \frac{1}{2} m (\frac{G M}{r}) = \frac{G M m}{2 r}$
$\Rightarrow K α \frac{1}{r} α \frac{1}{T^{2 / 3}} α T^{- 2 / 3}$

PHXI08:GRAVITATION

359719 An artificial satellite moving in a circular orbit around the earth has a total energy $E_{0}$ . Its potential energy is

1

- 2 E_{0}

2

E_{0}

3

- E_{0}

4

2 E_{0}

Explanation:

Potential energy $= 2 ($ total energy $) = 2 E_{0}$

PHXI08:GRAVITATION

359720 Match Column I with Column II. For a satellite in circular orbit, (where $M_{E}$ is the mass of the Earth, $m$ is mass of the satellite and $r$ is the radius of the orbit)
Column I
Column II
A
Kinetic energy
P
$- \frac{G M_{E} m}{2 r}$
B
Potential energy
Q
$\sqrt{\frac{G M_{E}}{r}}$
C
Total energy
R
$- \frac{G M_{E} m}{r}$
D
Orbital velocity
S
$\frac{G M_{E} m}{2 r}$

1 A - R, B - S, C - Q, D - P

2 A - Q, B - P, C - R, D - S

3 A - P, B - Q, C - S, D - R

4 A - S, B - R, C - P, D - Q

Explanation:

Kinetic energy $= \frac{G M_{E} m}{2 r}; A - S$
Potential energy $= - \frac{G M_{E} m}{r}; B - R$
Total energy $= - \frac{G M_{E} m}{2 r}; C - P$
Orbital velocity $= \sqrt{\frac{G M_{E}}{r}}; D - Q$
Option (4) is correct.

PHXI08:GRAVITATION

359721 The minimum energy required to launch a $m kg$ satellite from earth's surface in a circular orbit at an altitude of $2 R, R$ is the radius of earth, will be

1

3 mgR

2

\frac{5}{6} m g R

3

2 m g R

4

\frac{1}{5} m g R

Explanation:

Minimum required energy
$\begin{aligned} = T E_{f} - P E_{i} \\ = \frac{- G M m}{2 (R + 2 R)} - (\frac{- G M m}{R}) \\ = \frac{G M m}{R} (- \frac{1}{6} + 1) \\ = \frac{5}{6} \frac{G M m}{R} \\ = \frac{5}{6} \frac{g R^{2} \times m}{R} \\ = \frac{5}{6} m g R \end{aligned}$

PHXI08:GRAVITATION

359722 In a satellite if the time of revolution is $T$ , then kinetic energy is proportional to

1

T^{- 2 / 3}

2

\frac{1}{T}

3

\frac{1}{T^{2}}

4

\frac{1}{T^{3}}

Explanation:

The kinetic energy of the satellite is
$K = \frac{1}{2} m v_{0}^{2} = \frac{1}{2} m (\frac{G M}{r}) = \frac{G M m}{2 r}$
$\Rightarrow K α \frac{1}{r} α \frac{1}{T^{2 / 3}} α T^{- 2 / 3}$

PHXI08:GRAVITATION

359719 An artificial satellite moving in a circular orbit around the earth has a total energy $E_{0}$ . Its potential energy is

1

- 2 E_{0}

2

E_{0}

3

- E_{0}

4

2 E_{0}

Explanation:

Potential energy $= 2 ($ total energy $) = 2 E_{0}$

PHXI08:GRAVITATION

359720 Match Column I with Column II. For a satellite in circular orbit, (where $M_{E}$ is the mass of the Earth, $m$ is mass of the satellite and $r$ is the radius of the orbit)
Column I
Column II
A
Kinetic energy
P
$- \frac{G M_{E} m}{2 r}$
B
Potential energy
Q
$\sqrt{\frac{G M_{E}}{r}}$
C
Total energy
R
$- \frac{G M_{E} m}{r}$
D
Orbital velocity
S
$\frac{G M_{E} m}{2 r}$

1 A - R, B - S, C - Q, D - P

2 A - Q, B - P, C - R, D - S

3 A - P, B - Q, C - S, D - R

4 A - S, B - R, C - P, D - Q

Explanation:

Kinetic energy $= \frac{G M_{E} m}{2 r}; A - S$
Potential energy $= - \frac{G M_{E} m}{r}; B - R$
Total energy $= - \frac{G M_{E} m}{2 r}; C - P$
Orbital velocity $= \sqrt{\frac{G M_{E}}{r}}; D - Q$
Option (4) is correct.

PHXI08:GRAVITATION

359721 The minimum energy required to launch a $m kg$ satellite from earth's surface in a circular orbit at an altitude of $2 R, R$ is the radius of earth, will be

1

3 mgR

2

\frac{5}{6} m g R

3

2 m g R

4

\frac{1}{5} m g R

Explanation:

Minimum required energy
$\begin{aligned} = T E_{f} - P E_{i} \\ = \frac{- G M m}{2 (R + 2 R)} - (\frac{- G M m}{R}) \\ = \frac{G M m}{R} (- \frac{1}{6} + 1) \\ = \frac{5}{6} \frac{G M m}{R} \\ = \frac{5}{6} \frac{g R^{2} \times m}{R} \\ = \frac{5}{6} m g R \end{aligned}$

PHXI08:GRAVITATION

359722 In a satellite if the time of revolution is $T$ , then kinetic energy is proportional to

1

T^{- 2 / 3}

2

\frac{1}{T}

3

\frac{1}{T^{2}}

4

\frac{1}{T^{3}}

Explanation:

The kinetic energy of the satellite is
$K = \frac{1}{2} m v_{0}^{2} = \frac{1}{2} m (\frac{G M}{r}) = \frac{G M m}{2 r}$
$\Rightarrow K α \frac{1}{r} α \frac{1}{T^{2 / 3}} α T^{- 2 / 3}$

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